Use regular expressions:
require 'uri'
url = URI.parse('https://domain.com/////url/url2')
url.path.gsub! %r{/+}, '/'
p url.to_s
Pregunta
How to remove random excess of slashes from url or just validate it?
For example,
valid statements:
www.domain.com/url/url2
invalid statements:
https://domain.com/////url/url2
www.domain.com/url/////////url2
Thanks for help!
Solución
Use regular expressions:
require 'uri'
url = URI.parse('https://domain.com/////url/url2')
url.path.gsub! %r{/+}, '/'
p url.to_s
Otros consejos
this pattern do the job (with http(s) or not) :
"https://domain.com/////url/url2".gsub! %r{(?<!:)/+(?=/)}, ''
The other answers do not remove a trailing slash from the URL - which can be important for SEO purposes. There are many ways to do this, but for example:
require 'uri'
url = URI.parse('https://example.com/////url/url2/')
url.path.gsub! %r{/+}, '/'
url.path.sub! %r{/$}, ''
Or:
require 'uri'
url = URI.parse('https://example.com/////url/url2/')
url.path.squeeze!('/')
url.path.chomp!('/')
See: String#squeeze!
and String#chomp!
.